Asked by ues
Compare Linear Function 1 with Linear Function 2. Which one has the greater initial value? Choose 1 for Linear Function 1; choose 2 for Linear Function 2.
Linear Function 1: Taesha starts the year with $50 saved and then saves an additional $25 a week.
Linear Function 2: The relationship between the number of weeks Joshua has been saving (x) and his total savings (y) is given by y=37x+40
Linear Function ? has the greater initial value.
Linear Function 1: Taesha starts the year with $50 saved and then saves an additional $25 a week.
Linear Function 2: The relationship between the number of weeks Joshua has been saving (x) and his total savings (y) is given by y=37x+40
Linear Function ? has the greater initial value.
Answers
Answered by
GPT-4o mini
To compare the initial values of Linear Function 1 and Linear Function 2, we need to identify the initial value (the y-intercept) of each function.
**Linear Function 1**: Taesha starts the year with $50 saved and saves an additional $25 a week. The equation can be represented as:
\[ y = 25x + 50 \]
Here, the initial value (when \( x = 0 \)) is \( y = 50 \).
**Linear Function 2**: The function is given by:
\[ y = 37x + 40 \]
Here, the initial value (when \( x = 0 \)) is \( y = 40 \).
Now we can compare the initial values:
- Initial value of Linear Function 1 is $50.
- Initial value of Linear Function 2 is $40.
Since \( 50 > 40 \), **Linear Function 1 has the greater initial value**.
So, the answer is **1**.
**Linear Function 1**: Taesha starts the year with $50 saved and saves an additional $25 a week. The equation can be represented as:
\[ y = 25x + 50 \]
Here, the initial value (when \( x = 0 \)) is \( y = 50 \).
**Linear Function 2**: The function is given by:
\[ y = 37x + 40 \]
Here, the initial value (when \( x = 0 \)) is \( y = 40 \).
Now we can compare the initial values:
- Initial value of Linear Function 1 is $50.
- Initial value of Linear Function 2 is $40.
Since \( 50 > 40 \), **Linear Function 1 has the greater initial value**.
So, the answer is **1**.
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